SYLLABUS ‘Geo-technical Engineering: Properties of soil, classification, various tests and inter-relationships; Permeability & Seepage, Compressibili onsolidation and Shearin; stance, Earth pressure theories and stress distribution in sail; Soil exploration ~ planning & methods, Properties and uses of geo-syntnetics. Foundation Engineering: Types of foundations & selection criteria, bearing capacity, settlement analysis, design and testing of shallow & deep fou ndations; Slope stability analysis, Earthen embankments, Dams and Earth retaining structures: types, analysis and design, Principles of ground modifications —_ Origin of Soil and Soil Water Relationship Classification of Soil Soil Compaction Effective Stress, Capillary and Permeability ‘Seepage through Soil Vertical Stresses Compressibility and Consolidation Shear Strength of Soil Stability of Slopes: Earth Pressure and Retaining Walls. Shallow Foundation Deep Foundation Soil Exploration Geotextile 129 - 141 142 - 144 145 — 150 151 ~ 174 175 — 181 182 - 186 187 - 204 205 - 221 222 ~ 224 225 - 243 244-271 272 - 288 289 ~ 293 294 - 294 JES MASTER Publication Scanned with CamScanner: Ree 5) OF § D Son, CHAPTER \ Share RELATIONSHIp 1 f 1.75 gm/cc and G4: Soil isto be excavated from a borrow pit which has a density of Water con, ipacted so that water c ‘les is 2.7. The soil is com The specific gravity of soll partic! font content is rey a) fill, estimate (i) quantity of SOil to be excq, . For 1000 cum of soil in fil, Tene oa oa ott aco Also, determine the void ratiog oft at from had pe ie a 115 Marks Sy, Sol. (i) Siar w, y at LYS ws MN Borrow Pit For compacted fill, Compacted soit Dry density, =1.65 gmicm? 1650 kg/m? For 1000 cum of soi in fi, Weight of solid contents, Woois = 1850 % 1000 kg For borrow pit, Wooig (1+) Bulk density, Py met 750 kim? Water content, Y= 12% and (ii) Water content, For 1650 = 1000 kg of solids, ‘Amount of water in borrow pit sol, W, Amount of water in compacted til, Wa = 0.18 1650 x 1009 = 297000 kg ‘Amount of water to be added Wacta ( ~ 4) = 1650 x 1000 (0.18 ~ 0.12) 99,000 kg of water 0F 99 m? of wateruk oon ° 75 = 2Txt(1.42) tre, e, = 0.728 3 fill, compacted fil, a = Pw py densi a= Tre, 165 = 2221 2 tre; 2 values of liquid limit, plastic limit and shrinkage limit ofa soll were reported as below: at 2 608 5 Wp = 30% j Ws = 20% A ‘sample of this soil at liquid limit has a volume of 40 ce and its volume measured at shrinkage sieit WAS ‘23.5 cc, determine the specific gravity of the solids. What is its ‘shrinkage ratio and volumetric shrinkage? [14 Marks IES-1996] | ven: | got Data gi "a volume val of sample at shrinkage li Vol. of soil, since, beyond shrinkat eof sample at liquid limit, Vj,= 40 cm? 3.5 cm V = Vol. of solids (Vs) + Vol. of voids (Vy) .ge limit soil is completely saturated Vol. of water (Vw) Weight of water (Ww) Ww Water content (w)x Weight of solids (Ws) ~ tw wxW, Yw Vol. of voids (Vy) = v= Vet Ws = Vs Gs Yw V = Ve (1+ WG) V, (1+ Wi, Gs) Vg (14 Ws Gs) 140.665 = 740265 G, = 2.7049 " (40-235) 100 = —23:55—»100 = 1.755 (60-20) ) «100 = 70.213% ES MASTER Publication Scanned with CamScannerSol, 3 Thus, volume of soil solids, V, = 25 cm’ ere Volume of voids, V, = V~V, a2 = 08 Void ratio, © = y-* 25, M2?) aaa ~ Porosiiy.n = = 45 Q4: The void ratio and. 92% saturated, find Sol. Data given: Void ratio, ¢ Specific gravity, G Degree of saturation, § onal So 1, Pay, al the porosity and vot ay A samplor with @ youre ot wd eh ‘of water, What Is tho p : as: der, it alsplaces graduated cylinder, 2 = 45 om’ ; eee cee Joris displaced due tothe solid partes, f 1o th cylinder, wat ‘When soil is powered int 2.73 and 2.7 respectively, ty of a sample of clay are te ho baboons the dry density and the water content, it, What would be the water content for complete saturation, Bulk density, », = Water content = We know that es a w Dry density ple. When the soi ith a soll samy Ned wl 4 oF, the vold ratio remaining the 19.118 wxar found to h: % and aunt 2 Bil. Laboratory ea ane tama ne cnet Sanda minimum and maximum Wold ratios respectively for denset and looser ‘States. Calculate the degree! Saturation and the relative density, Assume G = 2 (10 Marks S-200) Sol. Data given Natural moisture content, = 15% Unit weight, y= 16.84 kNim® Speciic gravity, G = 265 Maximum void ati, e,., = gas Minimum void ratio, = 05 But unit weight, We know that, Se y= ese), Te “G Scanned with CamScannercrileni’' eee ERING SSS MECH 5, D ENGINE "SOIL. MECHANICS & FOUNDATION ENGINE! m 4 = SMW . te . 1984 = 295 (140.18)x9.81 ite = e = 0.5868 using, Se = we = S * 0.5868 = 0.15 x 2.65 = S = 06774 = 67.74% Relative density, (D,) = 0.85-0.5868 mn 085-05 = 0.7519 or 75.19% 6: Earth is required to be excavated from borrow pits for building an embankment of height 6 m, top width 2m and side slopes 1 : 1. The unit welght of undisturbed soil in wet condition is 18 kN/m? and its natural water content is 8 percent. The dry density required in the embankment is 20 kN/m? with a water content of 10%, The specific gravity of soil solids is 2.70. Estimate the quantity of earth required to be excavated in the borrow area to construct one metre length of the embankment. If each truck has a capacity to carry 80 KN per trip, what is the number of truck loads required per metre length of embankment? What are the values of porosity and degree of saturation on the embankment? [15 Marks IES-2003] Sol. le— 6m —re 2m >4— 6m —F We will use subscript 1 for borrow pit and 2 for embankment respectively. Bulk unit weight of borrow pit soil, Yh, = 18 kNim? Water content of borrow pit soil, w, = 8%. Dry density of embankment soil, gp = 20 KNim? _ Water content of embankment soil, wy = 10%. = 27 Specific gravity of soil solids, G, = 1 ean J (a4 1g)x6 1 = 48m Volume of embankment per unit of length, = 2%(2+14) bi t - Calculations for finding volume of solids (V,) in *m length of embankment Wy Mesion #O> tn Voot 7 Vea) MASTER Publication f Scanned with CamScanner1 Mite, aril 1995-202) ry | onive vated from borrow pit ars iy ot ent obo exceed ns for finding quan Calculatlo t soll, Y4y Bulk unit woight of borrow pit Gre Youn(*™ ne ‘yo . : 36.244x(14-0.08) - 57.6 mys 2.79.81) hea 2 — io m 1 bo excavated from borrow pit area. Thus, 57.6 m* of earth is required 1 a Weight of 57.6m! of borrow pit soll, W, = 57: = 1036.8 _ 42.96 Number of truck loads required = 80 = 1036.8 kN f embankment, Thus 13 number of truck loads are required per metre length of Calculations for porosity (n) and degree of saturation (S) of embankment soil om Ye * Tre, = 2 = e = — = 0.245 we "= Tre, 1+0.324 Also, eS = WG = 0.324 $= 01% 27 = S, = 0.8324 = 83.24% O7: Explain and discuss tho use of Liquidity Index, Activity number, Thixotropy and Sensitivity ofa. [10 Marks ies) Sol. Liquidity Index (,) + Liquidity index is defined as the ratio of the ‘differen Plastic limit (w,)' to the plasticity index ie. 'ce between natural water content (of sola @) Activity Number (A,) * Activity number isthe ratio of the plastic ioe finer than 2 microns) N=IY index (I,) 0 clay fraction (e., percentage by mass oP Scanned with CamScanner&: ee — mex p reer | an SOIL. MECHANICS & FOUNDATION ENGINEER'S mineral Kaoiite are least active (A 1), whoreas tho sols containing tho mineral lite are moderately active (A = 1 to 2) + Depending upon activity, the soils aro classified into three typos: cvsEngineering S.No. Activity Soll type A< 0.75 Inactive 2 A= 0.7510 1.25 Normal 3 A> 1.25 Active {@) Thixotropy: + Itis that property of soil due to which loss of strength (shear strength) on remoulding can be regained if left undisturbed for some time. + Increase in strength with passage of time is due to tendency of clay soil to regain their chemical equilibrium with the reorientation of water molecule in adsorbed layer. 1+ Thixotropy of soils is of great practical importance in soil engineering. For example, if pile is driven into soil, it disturbs the soil. Frictional resistance of driven pile in clay soil immediately after installation will be much less than its value after a month. (4) Sensitivity (S): + Sensitivity is defined as the ratio of the unconfined compressive strength of an undisturbed ‘soil sample to the unconfined compressive strength of the specimen of the same soil after remoulding it at an unaltered moisture content. 4, (undistrubed) S, = “q, (remoulded) + Ithelps in measuring the effect of remoulding of soil on its strength. Q8: An oven dry soil sample of volume 225 cm’ weighs 390 g. If the specific gravity is 2.72, determine the void ratio and shrinkage limit. What will be the water content which will fully saturate the sample and cause an increase in volume equal to 8% of the original dry volume? [10 Marks IES-2011] Sol. Volume of dry sample of soil, V = 225 cm? Weight of dry soil or soil solids, W., = 390 gm Specific gravity of soil, G = 2.72 (i) We know that, Dry density of soil, 74 = tre V - 27a! we neato > e = 0.569 At shrinkage limit, soil is fully saturated i.e. S = 1 Now eS = WG * 0.569 * 1 = wx 2.72 S w = 20.93 % Thus shrinkage limit of soll, __W, = 20.93% IES MASTER Publication Scanned with CamScannerN Conventional Solved Papers 1995-2021 7 = 2 <225= 18m " Increment in volume, AV = t | 00 X Ve = OM. = Vs = — dry sample, Vy ive Vol. originally occupied by voids in dry 0.509% 228 = 1.618 om? Ww * 10569 = ids i to be fully saturated, the volume occupled by voids in dry Samp mH ince, the soil is given " t roca volume (AV) will be occupied Pair ey ‘ex Volume occupied by water finally, Vy einen 99.618 cm? x 3 Vu * Pw = 99.618 x tgmicm = Weight of water, Seacian Wa, 99.618 49 Water content, w = 3g =F = 25.54% Ans. @9: Whatao YOU understand by “index Properties of soil"? Explain and. list the ‘Properties “Under dite cat ies, [4 Marks KES. Sol. Index properties of soi fe en Properties which help to access the Engineering behaviour ofa sa Me 8SSist in determining its classification accurately, The index Properties for Coarse and fir Type of soil Index property Coarse soit Particle size distribu Fine soit Atterberg’s limit and consistency, INg gradation and uni °f Various sizes of particles resent in a Hormity of soil, 7 ams, embankment, ters en his knowledge helps in construction of es? (2) Grain shape: + Its defined as; * Itrepresents degree of «, Oarseness, + Higher relative density means higher shear strengy aR ep lenseness or g i. natural Seposits of coarse grained 5 th and low. Compressibility and vice-ves? Scanned with CamScannerAttorbera’s Limit and Consistency; «Consistency reprosents the relative oa: + Consistency is related to water conten changes. 'se with which A soil can be deformed. ''1.0., how with change in water content, the consistency of soll . pe {he consistoncy in four stages, Behaviour of sol Ia eliforent in diferent stages. ) Solid stage ——_(b) Semi solid stay id @ We shape yO Stage (6) Plastic slago (A) liquid stage \W, plastic iit water content 1W/= liquid lim water content | Water content» « Itcan be seen that soil becomes more and more ductile as its water content increases. 0-40: soil sample has a porosity of 40%, The specific gravity of solids is 2.7. Calculate the (i) void ratio (i) dry density (ii) unit weight if the soil is 50% saturated and (iv) unit weight if the soil is completely saturated. [5 Marks 1ES-2012] Sol, Porosity of soil, n= 40% | Specitic gravity of soll solids, G = 27 E ‘a iatis’s =e t's 04 | Void ratio = 72 = ==> = 0.687 sty, 7, = Gt = 27*9.81 _ 5 ) Dry density, yg = pee = TAEST = 16.002 kNim )_Unit weight of soil if it is 60% saturated, | _ GtS. - | é Mooi tee Mon pice = 17.854 N/m? |) Unit weight of soit if it is completely saturated, _ G+Se n° tre ™ 2.7 +1x0.667 s o——— «981 7 n T0687 “°° a th = 19.816 kNim? i The mass of saturated soil sample is 150 gm and its mass when oven dried is 90 gm, find the water Content. Suppose that the sample, used for triaxial test, has a diameter of 38 mm and the height of 76 mm, find the void rati [4 Marks IES-2013] = © word ratio. Scanned with CamScannerSaturated soll sample Oven dried soit Data given, Nas o saturated soil sample, W = 150 gm Mass of soll when over dried, W, = 90 gm Diameter of the soil sample, d = 38 mm Height of the soil sample, h = 76 mm Mass of water present in saturated soil sample, W, = W-W, = 150 90 = 60 gm. Water content, w " Tie 100 = 82,499 = 66.67% Rg Volume of the soil sample, v = grdxh eens = 563.82x7.6 4 = 86.193 cms When soils fuly saturated, entre old space is occupied by water Wy 60, ie. Volume of voids (v,) = y, = Pe “TgmTSE = 60 em? Volume of solids (v,) = y_y 86.193 ~ 69 = 26.193 cm3 Void ratio, @ << O12: A liquid limit test Conducted on a soit sam ple in the cup device gave the following Number of blows 10 19 23 a Water content (%) 60.00 45.29 39.80 36.50 Two determinations for the Plastic limit gave water Content of 20.30% and 20.80% Determine @ The liquid limit and plastic limit (i) The plasticity index (il) The liquidity index i¢ the natural w; i) The void ratio *t2r Content is 27.40% nd 3t the liquid timit, ix 1 © SP. gravity, You expect brittle failure? % failure, would Scanned with CamScanner{qu imi Is tho water content at vhich 26 Wo wil hava to do intorpotation blows clos the grove, w)-30.8 1,398 1.302 i . Thu 305-300 " 4431-4.902 ‘. wy © 38070% 20.3) 20.8 Plastic Umit, w, = “SADE. 20.56% 0 Plastic indox, |, = w,~w, = 96.08 ~ 20.55 = 1753 Liquidity, indox |, = /=Me.. 274 -20.55. @ it 8.08 - 20.55 (iv) At liquid limit, degree of saturation, s=4 We know that, 0S = WG 2 1 = 38.08 * 2.7 Void ratio, © = 1.028 The liquidity index (I) of soil lies between 0.25 and 0.5. ‘Thus the consistency of soll is medium stiff, Hence it will undergo plastic failure. Had the liquidity index been loss than O, we would have expected a brittle failure. ‘A proposed earthen dam will have a volume of 5000000 of compacted soll. The soil is to be taken from a borrow pit and will be compact to a void ratio of 0.8. The void ratio of soil in the borrow pit Is 115. Estimato tho volume of the soll that must be exacted from the borrow pit for the construction of the above dam. [4 Marks IES-2015] Solid weight will be same (Yas = (tao ‘fa, = dry unit woight of borrow pits = Vol. of borrow pit soil ‘ta, = dty unit weight of compacted sol = vol. of compacted soil on IES MASTER Publication Scanned with CamScanneron i has a vold ratio of 0.70, degree of saturation 50% and G, =2.7. Find the Wate, t sol .70, F e¢ bulk density and dry density. By how much can the water content be Increaseg Maye iy, fe Mag, % | Sol. Given that Pent 8 = 05 Gs = 27 Find 1. Yes Yay e So, Porosity (n) = <> n= 0.412] Gw = Se => 27%w= 05% 07 = Water content = w= 0.1296 or w= 12.9694 toa! = SO+w) tw = 2.7 (140, 1296) 9.81 the 1407 Vary = G3) we For ‘Saturating the Soil at current Void ratio. a S41 Gw = Se = 1x07 0727 0.2592 0.2582 - 0.1296 | canned with CamScanner ———,oe You = 20 kNim? Gyw = So m ,*0.23 = tre © = 0296, ‘. = GtSehy Ysat tte ie 29 = (Gs41%0.296,)x10 1*0.296, & G, = 2817 Specific gravity of cay solid (G,) = 2617 Note The data of mass spectc gravity of oven ay sam (W,) as IES MASTER Publication Scanned with CamScanner